It's pretty simple.

The formal definition of velocity is the derivative of position with respect to time. So in one dimension:

$v = \frac{dx}{dt}$

And therefore

$x(t) = \int_{0}^{t}v(t')dt'$

If you assume $v(t)$ is constant, then that equation becomes $x = vt$ (hence the kinematic equation). If $v(t)$ isn't constant, then $x=vt$ is almost certainly not correct.

It's pretty simple.

The formal definition of velocity is the derivative of position with respect to time. So in one dimension:

$v = \frac{dx}{dt}$

And therefore

$x(t) = \int_{0}^{t}v(t')dt'$

If you assume $v(t)$ is constant, then that equation becomes $x = vt$ (hence the kinematic equation). If $v(t)$ isn't constant, then $x=vt$ is almost certainly not correct.

Edit: As @hft pointed out, the more central assumption to kinematics is that $a(t)$ is constant, so that:

$v(t) = \int_{0}^{t}a(t')dt' = at$

Now if we substitute $at$ in for $v(t)$, we get:

$x(t) = \int_{0}^{t}v(t')dt' = \int_{0}^{t}at' + v_0dt' = \frac{1}{2}at^2 + v_0t$

Which should look like the form we know and love.

It's pretty simple.

The formal definition of velocity is the derivative of position with respect to time. So in one dimension:

$v = \frac{dx}{dt}$

And therefore

$x(t) = \int_{t_1}^{t_2}v(t)dt$

If you assume $v(t)$ is constant, then that equation becomes $x = vt$ (hence the kinematic equation). If $v(t)$ isn't constant, then $x=vt$ is almost certainly not correct.

It's pretty simple.

The formal definition of velocity is the derivative of position with respect to time. So in one dimension:

$v = \frac{dx}{dt}$

And therefore

$x(t) = \int_{0}^{t}v(t')dt'$

If you assume $v(t)$ is constant, then that equation becomes $x = vt$ (hence the kinematic equation). If $v(t)$ isn't constant, then $x=vt$ is almost certainly not correct.